HDU 1024 Max Sum Plus Plus

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13897    Accepted Submission(s): 4569

Problem Description

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

Given a consecutive number sequence S₁, S₂, S₃, S₄ ... S_x, ... S_n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S_x ≤ 32767). We define a function sum(i, j) = S_i + ... + S_j (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + sum(i_m, j_m) maximal (i_x ≤ i_y ≤ j_x or i_x ≤ j_y ≤ j_x is not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i_x, j_x)(1 ≤ x ≤ m) instead. ^_^

 

Input

Each test case will begin with two integers m and n, followed by n integers S₁, S₂, S₃ ... S_n.
Process to the end of file.

 

Output

Output the maximal summation described above in one line.

 

Sample Input

1 3 1 2 3

2 6 -1 4 -2 3 -2 3

 

Sample Output

6

8

Hint

Huge input, scanf and dynamic programming is recommended.

 

Author

JGShining（极光炫影）

 

思路：DP，使用二维数组会超空间，所以我选择滚动数组

 

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;
int n,m;
int map[1000010];
long long dp[2][100000];
long long maxn[100000];
int the_last_sum;
int max(int x,int y)
{
    return x > y ? x : y;
}
int main()
{
    while(~scanf("%d%d",&m,&n))
    {
        memset(map,0,sizeof(map));
        for(int i = 1;i <= m;i ++)
            maxn[i] = -999999999;
        for(int i = 0;i <= 2;i ++)
            for(int j = 1;j <= m;j ++)
               dp[i][j] = -999999999;
        for(int i = 1;i <= n;i ++)
            scanf("%d",&map[i]);
        dp[1][1] = map[1];
        for(int i = 1;i <= n;i ++)
            for(int j = m;j >= 1;j --)
            {
                dp[i % 2][j] = max(dp[(i - 1) % 2][j] + map[i],maxn[j - 1] + map[i]);
                if(dp[i % 2][j] > maxn[j])
                     maxn[j] = dp[i % 2][j];
            }
       printf("%lld
",maxn[m]);
    }
    return 0;
}